1. Proof Must Have Statement of what is to be proven.
"Proof:" to indicate where the proof starts
Clear indication of flow
Clear indication of reason for each step
Careful notation, completeness and order
Clear indication of the conclusion

2. Number Theory - Ch 3 Definitions
Z --- integers
Q - rational numbers (quotients of integers)
rQ  a,bZ, (r = a/b) ^ (b  0)
Irrational = not rational
R --- real numbers
superscript of positive portion only
superscript of negative portion only
other superscripts: Zeven, Zodd , Q>5
"closure" of these sets for an operation
Z closed under what operations?

3. Integer Definitions even integer odd integer prime integer (Z>1)
n Zeven  k  Z, n = 2k
odd integer
n  Zodd  k  Z, n = 2k+1
prime integer (Z>1)
n Zprime  r,sZ+, (n=r*s) (r=1)v(s=1)
composite integer (Z>1)
n  Zcomposite  r,sZ+, n=r*s ^(r1)^(s1)

4. Constructive Proof of Existence
If we want to prove:
nZeven, p,q, r,sZprime n = p+q ^ n = r+s
^pr^ ps^ qr^ qs
let n=10
n Zeven by definition of even
Let p = 5 and the q = 5
p,q Zprime by definition of prime
10 = 5+5
Let r = 3 and s = 7
r,s Zprime by definition of prime
10 = 3+7
and all of the inequalities hold
Make sure to mention the appendix that has rules of algebra.

5. Methods of Proving Universally Quantified Statements
Method of Exhaustion
prove for each and every member of the domain
rZ+ where 23<r<29   p,q Z+ (r = p*q)^(p<=q)
Generalizing from the "generic particular"
suppose x is a particular but arbitrarily chosen element of the domain
show that x satisfies the property
i.e. rZ, r Zeven  r2 Zeven
Exhaustion: Let r be an arbitrary member of Z
24 = 4* Assume r Zeven
25 = 5*5 exists k in evens such that r = 2k (def of even)
26 = 2* when squaring both sides of r = 2k, we get r2 = 4k2
27 = 3*9 by factoring out a 2 we get r2 = 2(2k2)
28=2* k2 is even because 2k2 is an integer by closure
therefore r2 is even
r in Zeven -> r2 in Zeven

6. Examples of Generalizing from the "Generic Particular"
The product of any two odd integers is also odd.
m,n Z, [(m  Zodd  n Zodd ) m*n Zodd ]
The product of any two rationals is also rational.
m,nQ, m*n Q

7. Disproof by Counter Example
 rZ, r2Z+  rZ+
Counter Example: r2= 9 ^ r = -3
r2Z+ since 9 Z+ so the antecedent is true
but rZ+ since -3Z+ so the consequent is false
this means the implication is false for r = -3 so this is a valid counter example
When a counter example is given you must always justify that it is a valid counter example by showing the algebra (or other interpretation needed) to support your claim

8. Division definitions d | n  kZ, n = d*k n is divisible by d
n is a multiple of d
d is a divisor of n
d divides n
standard factored form
n = p1e1 * p2e2 * p3e3 * …* pkek

9. Proof using the Contrapositive
For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d.

10. Proof using the Contrapositive
For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d.
n,d,cZ+, ndc  nd

11. Proof using the Contrapositive
For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d.
n,d,cZ+, ndc  nd
n,d,cZ+, n|d  n|dc
proof:
let n, d and c be arbitrary in Z+
assume n|d
there exists a k in Z such that d=nk by def of divides
multiplying both sides by c gives dc = nkc
kc in Z by closure of Z in multiplication
n|dc by definition of divides
n|d -> n|dc by closing conditional world w/out contra
forall n,d,c in Z+, n|d -> n|dc by generalizing from GP

12. more integer definitions
div and mod operators
n div d --- integer quotient for
n mod d --- integer remainder for
(n div d = q) ^ (n mod d = r)  n = d*q+r
where n Z0, d Z+, r Z, q Z, 0 r<d
relating “mod” to “divides”
d|n  0 = n mod d
 0 d n
definition of equivalence in a mod
x d y  d|(x-y) [note: their remainders are equal]
sometimes written as x  y mod d meaning (x  y) mod d
examples of equivalence in a modulous 122,12 mod 10 & 10,13 mod 3

13. Quotient Remainder Theorem
nZ dZ+ q,rZ
(n=dq+r) ^ (0  r < d)
Proving definition of equiv in a mod by using the quotient remainder theorem
This means
prove that if [m d n], then [d|(n-m)]
where m,nZ and dZ+
Using what you know about mod operator from programming
that the remainder is the same
exists k,l,r in Z such that n = dk+r and m = dl+r
n=dk+r n-dk = r
m = dl + r m-dl = r
n-dk = m-dl
n-m = dk-dl
n-m = d(k-l)
k-l in Z by closure
so d| (n-m) by definition of divides

14. Proofs using this definition
mZ+ a,bZ
a m b   k Z a=b+km
mZ+ a,b,c,dZ
a m b ^ c m d  a+c m b+d
# #2
m in Z+ and a,b in Z are GP m in Z+, a,b,c,d in Z are GP
Assume a m-equiv b assume a m-equiv b and c m-equiv d
m|(a-b) by def of equiv in a mod m | (a-b) and m|(c-d) by def of equiv in a mod
exists k in Z such that (a-b) = mk exists k in Z such that a-b = km
a = b + km by algebra exists x in Z such that c-d = xm
by def of divides
Assume exists k in Z a=b+mk
a-b = mk (a-b)+(c-d) = km+xm by adding equals to equals
Since k is an integer, (a+c)-(b+d) = (k+x)m by algebra
m|(a-b) by def of divides Since k+x in Z by closure of Z in addition
a m-equiv b by def of equiv in a mod m| (a+c)-(b+d) by definition of divides
(a+c) m-equiv (b+d) by def of equiv in a mod

15. Proof by Division into Cases
nZ 3n  n2 3 1
# #2
forall n in Z [n/2] is Let n be arbitrary in Z
n/2 if n is even We know by the quotient remainder theorem that
or (n-1/2) if n is odd there exists a q and r such that
proof: n = 3q+0 or n = 3q+1 or n=3q+2
Case 1:Assume n is even
exists k in Z n = 2k Taking these three cases:
by def of even Case 1: (assume: n=3q+0)
exists x in Z where x = floor n/2 n=3q and 3|n which means the antecedent is false
x <= n/2 < x and the implication is true
x <= 2k/2 < x+1
x <= k < x+1
x = k
Case 2: (assume n = 3q+1)
Case 2: Assume n is odd since n=3q+1, n^2 (3q+1)^2=9q^2+6q+1=3(3q^2+2q)+1
exists w in Z n = 2w n^2-1 = 3(3q^2+2q)
by def of odd Since 2q^2+2q is in Z by closure of Z in * and +
exists x in Z when x = floor n/ | n^2 –1 by the definition of divides
x <= n/2<x and n^2 3-equiv to 1 by def of equiv in a mod
x <=(2w+1)/2<x+1
x<= w+1/2 < x Case 3: (assume n = 3q+2)
x <= w < x+1 since n=3q+2,n^2=(3q+2)^2=9q^2+18q+4=3(3q^2+6q+1)+1
x = w n^2 – 1 = 3(3q^2+6q+1)
n = 2w q^2+6q+1 is in Z by closure of Z in * and +
n-1 = 2w 3|n^2-1 by definition of divides
(n-1)/2 = w n^2 equiv to 1 by def of equiv in a mod
(n-1)/2 = x implication true in all three cases

16. Floor and Ceiling Definitions
n is the floor of x where xR ^ n Z
x = n  n  x < n+1
n is the ceiling of x where xR ^ n Z
x = n  n-1 < x  n

17. Floor/Ceiling Proofs x,yR x+y = x + y xR yZ x+y = x +y
#1 disproof by counter example #2 proof by GP
let x = 2.5 and let y = let x be arbitrary in R and let y be arbitrary in Z
then floor(x) = 2 and floor(y) =2 by the definition of floor of x
and floor(x)+floor(y) = exists n in Z such that n <= x<n+1
adding y to all sides of the inequality we get
x+y = n+y <= x+y < n+y+1
then floor(x+y) = by closure n+y and n+1+y are in Z
so this is the definition of the floor of x+y
which are not equal

18. Proof by Division into Cases (again)
The floor of (n/2) is either
a) n/2 when n is even
or b) (n-1)/2 when n is odd
# #2
forall n in Z [n/2] is Let n be arbitrary in Z
n/2 if n is even We know by the quotient remainder theorem that
or (n-1/2) if n is odd there exists a q and r such that
proof: n = 3q+0 or n = 3q+1 or n=3q+2
Case 1:Assume n is even
exists k in Z n = 2k Taking these three cases:
by def of even Case 1: (assume: n=3q+0)
exists x in Z where x = floor n/2 n=3q and 3|n which means the antecedent is false
x <= n/2 < x and the implication is true
x <= 2k/2 < x+1
x <= k < x+1
x = k
Case 2: (assume n = 3q+1)
Case 2: Assume n is odd since n=3q+1, n^2 (3q+1)^2=9q^2+6q+1=3(3q^2+2q)+1
exists w in Z n = 2w n^2-1 = 3(3q^2+2q)
by def of odd Since 2q^2+2q is in Z by closure of Z in * and +
exists x in Z when x = floor n/ | n^2 –1 by the definition of divides
x <= n/2<x and n^2 3-equiv to 1 by def of equiv in a mod
x <=(2w+1)/2<x+1
x<= w+1/2 < x Case 3: (assume n = 3q+2)
x <= w < x+1 since n=3q+2,n^2=(3q+2)^2=9q^2+18q+4=3(3q^2+6q+1)+1
x = w n^2 – 1 = 3(3q^2+6q+1)
n = 2w q^2+6q+1 is in Z by closure of Z in * and +
n-1 = 2w 3|n^2-1 by definition of divides
(n-1)/2 = w n^2 equiv to 1 by def of equiv in a mod
(n-1)/2 = x implication true in all three cases

19. Prime Factored Form kZ, p1,p2,…pkZprime, e1,e2,…ekZ+,
n = p1e1 * p2e2 * p3e3 * …* pkek
Unique Factorization Theorem (Theorem 3.3.3)
given any integer n>1
kZ, p1,p2,…pkZprime, e1,e2,…ekZ+,
where the p’s are distinct and any other expression of n is identical to this except maybe in the order of the factors
Standard Factored Form
pi < pi+1
mZ,8*7*6*5*4*3*2*m=17*16*15*14*13*12*11*10
Does 17|m ??

20. Steps Toward Proving the Unique Factorization Theorem
Every integer greater than or equal to 2 has at least one prime that divides it
For all integers greater than 1,
if a|b, then a (b+1)
There are an infinite number of primes
#1 Proof by div into cases #2 let a and be be arbitrary in Z where a > 1
case 1: n is prime assume a|b
it has a prime that divides it exists k in Z such that b = ak
(itself) assume a | b+1
case 2: n is not prime exists m in Z such that b+1=am
we need chapter 4 to prove this ak+1 = am by substitution
1 = am-ak=a(m-k)
1/a = m-k
1/a is an integer only when a = 1
but m-k is an integer by closure Contradiction
a not| b+1
#3 assume there are a finite number of primes
lets say the number of primes is m
We can take that list of primes and find their product
p1*p2*p3*…*pm = k
k is an integer by closure of integers in multiplication
Lets look at k+1
since p1|k, p1 not| k+1
since p2|k, p2 not|k+1 …
but by the first proof on this page there must be an integer that divides k+1
that isn’t in this list of m integers contradiction
so there must be a finite number of primes

21. Using the Unique Factorization Theorem
Prove that the
Prove:
aZ+qZprime q|a2  q |a
#1 Assume the cuberoot(3) is rational
there exists a,b in Z such that cuberoot(3) = a/b and b neq 0
cubroot(3) = a/b
b(cuberoot(3)) = a
3*b^2 = a^2
by the prime factorization theorem
a can be prime factorized to p1^e1*p2^e2*…px^ex
b can be prime factorized to q1^d1*q2^d2*…qy^dy
a^2 = p1^2e1*p2^2e2*…px^2ex
b^2 = q1^2d1*q2^2d2*…qy^2dy
By the unique prime factorization theorem
one of the p’s must be 3
factor all of the three’s out
a^2 = 3^2ei(a1^2) where a1 is the left over product of primes
and it is important to know that 3 not| a1
3*b^2=3^2ei(a1^2)
cancel a 3 from both sides and we get b^2=3^(2ei-1)(a1^2)
Since ei >=1, there is still a 3 left on the right
so one of the q’s must be a 3 factor all of the three’s out
to get b^2 = 3^2dj(b1^2) where b1 is the left over product of primes
and it is important to know that 3 not|b1
by substitution we get 3^2dj(b1^2) = 3^(2ei-1)(a1^2)
by the unique prime factorization theorem 2dj = 2ei-1
contradiction since one is odd the other is even
Prove: a,b,cZ+pZprime
[(a2 = b3 = c) ^ (p|c)] (p6 | c)
(a^2=b^3=c) and (p|c)
since p|c – exists k in Z such that c=pk
Since a^2 = c a can be prime factorized and a^2 has 2 p’s
since b^3 = c and b can be prime factorized b^3 has 3 p’s
since c must have a multiple of 2 p’s and a multiple of 3 p’s it must have a multiple of 6 p’s

22. Summary of Proof Methods
Constructive Proof of Existence
Proof by Exhaustion
Proof by Generalizing from the Generic Particular
Proof by Contraposition
Proof by Contradiction
Proof by Division into Cases

23. Errors in Proofs Arguing from example for universal proof.
Misuse of Variables
Jumping to the Conclusion (missing steps)
Begging the Question
Using "if" about something that is known